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The equation for the combustion of glucose is: C6H12O6(s)+6O26CO2(g)+6H2O . How many grams of H2O will be produced when 8.064g of glucose is burned?

Answer
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Hint :Combustion reaction: A type of chemical reaction in which a compound is heated in the presence of air to produce carbon dioxide and water. All combustion reactions are exothermic reactions as an appreciable amount of heat is released after the reaction.

Complete Step By Step Answer:
For finding mass of water produced, we first need to find out the number of moles of glucose burned. We know that, number of moles of any compound is equal to the ratio of mass used to the molar mass of the compound. Therefore, number of moles of glucose burned during the reaction are as follows:
 n=givenmassmolarmass
Given mass of glucose i.e., mass of glucose burned during the reaction =8.064g
Molar mass of glucose =180.16gmol1
  number of moles of glucose n=8.064180.160.045moles
The chemical reaction for combustion of glucose is given as follows:
 C6H12O6(s)+6O26CO2(g)+6H2O
As per the given reaction,
 1 mole of glucose reacts to produce 6 moles of H2O
Therefore, 0.045 moles of glucose will react to produce 6×0.045=0.27 moles of H2O
Now, mass of water produced can be calculated using mole-mass relation which is as follows:
Number of moles of H2O produced =massmolarmass
Molar mass of H2O=18gmol1
Substituting values in the equation,
 0.27=mass18
 mass=18×0.27
 mass=4.86g
Hence, 4.86g of H2O is produced after the combustion of 8.064g glucose.

Note :
Combustion reaction is a type of oxidation reaction because of addition of oxygen atoms to the given compound. It is important to note that all combustion reactions are oxidation reactions but all oxidation reactions are not combustion reactions.
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